Question 226228
The equation should be {{{(sqrt(n)+sqrt(n-1))(sqrt(n)-sqrt(n-1))=1}}}. Make sure that you have the correct equation.



a) Let's show that {{{(sqrt(n)+sqrt(n-1))(sqrt(n)-sqrt(n-1))=1}}} is true for the values {{{n=1}}} and {{{n=2}}}



Let's show the equation is true for n=1:



{{{(sqrt(n)+sqrt(n-1))(sqrt(n)-sqrt(n-1))=1}}} Start with the given equation.



{{{(sqrt(1)+sqrt(1-1))(sqrt(1)-sqrt(1-1))=1}}} Plug in {{{n=1}}}



{{{(sqrt(1)+sqrt(0))(sqrt(1)-sqrt(0))=1}}} Combine like terms.



{{{(1+0)(1-0)=1}}} Take the square root of 1 to get 1. Take the square root of 0 to get 0.



{{{(1)(1)=1}}} Combine like terms.



{{{1=1}}} Multiply.



So {{{(sqrt(n)+sqrt(n-1))(sqrt(n)-sqrt(n-1))=1}}} is true when {{{n=1}}}



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Let's show the equation is true for n=2:



{{{(sqrt(n)+sqrt(n-1))(sqrt(n)-sqrt(n-1))=1}}} Start with the given equation.



{{{(sqrt(2)+sqrt(2-1))(sqrt(2)-sqrt(2-1))=1}}} Plug in {{{n=1}}}



{{{(sqrt(2)+sqrt(1))(sqrt(2)-sqrt(1))=1}}} Subtract



{{{(sqrt(2)+1)(sqrt(2)-1)=1}}} Take the square root of 1 to get 1.



{{{sqrt(2)*sqrt(2)-sqrt(2)+sqrt(2)+1(-1)=1}}} FOIL



{{{2-sqrt(2)+sqrt(2)-1=1}}} Multiply



{{{1=1}}} Combine like terms.



So {{{(sqrt(n)+sqrt(n-1))(sqrt(n)-sqrt(n-1))=1}}} is true when {{{n=2}}}



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b)



Now let's show that {{{(sqrt(n)+sqrt(n-1))(sqrt(n)+sqrt(n-1))=1}}} is true for any value of 'n' such that {{{n>=1}}}



{{{(sqrt(n)+sqrt(n-1))(sqrt(n)+sqrt(n-1))=1}}} Start with the given equation.



Let {{{x=sqrt(n)}}} and {{{y=sqrt(n-1)}}}



{{{(x+y)(x-y)=1}}} Replace each {{{sqrt(n)}}} with 'x' and replace each {{{sqrt(n-1)}}} with 'y'



{{{x^2-xy+xy-y^2=1}}} FOIL



{{{x^2-y^2=1}}} Combine like terms.



{{{(sqrt(n))^2-(sqrt(n-1))^2=1}}} Plug in {{{x=sqrt(n)}}} and {{{y=sqrt(n-1)}}}



{{{n-(n-1)=1}}} Square each square root to eliminate them (note: because {{{n>=1}}}, this means that {{{n-1>=0}}} which means that each radicand is positive. So we can avoid absolute values here).



{{{n-n+1=1}}} Distribute



{{{1=1}}} Combine like terms.



So this shows that {{{(sqrt(n)+sqrt(n-1))(sqrt(n)+sqrt(n-1))=1}}} is true for any value of 'n' such that {{{n>=1}}}